and that the

D and E are right, and AC is common to the two ADC, AEC, .. (E. 26. 1.) the EAC=<DAC: Again, since (constr. and E. 5. 1. cor.) the ACB= ABC, and (constr.) the at D are right angles, and that AC AB, ... (E. 26. 1.) the DAC = DAB: But it was shewn that the 4 EAC

DAC: .. ¿ EAC= 2 DAC= 2 DAB; i. e. the right XAY is trisected by AC and AD.

PROP. XXI.

31. PROBLEM. Hence, to trisect a given rectilineal angle, which is the half, or the quarter, or the eighth part, and so on, of a right angle.

First, let the given YAZ, be the half of a

right, and let it be required to trisect it.

1

Draw (E. 11. 1.) from A, AX AY; trisect (S. 18.1.) the right XAY; then (S. 1. 1.) trisect the YAZ, which is the half of the YAX. But, if the given be the quarter of a right angle, its double may be trisected by the former case; and... the given itself may be trisected

by (S. 1. 1.)

And, by following the same method, it is evident that an may be trisected, which is the eighth part, or the sixteenth part, and so on, of a right angle.

PROP. XXII.

32. PROBLEM. In the hypotenuse of a right-angled triangle, to find a point, the perpendicular distance of which from one of the sides, shall be equal to the segment of the hypotenuse between the point and the other side.

Let ABC be a right-angled A, right-angled

at C: It is required to find a point in the hypo

tenuse AB, the perpendicular distance of which from one of the sides, as AC, shall be equal to the segment of the hypotenuse between that point, and BC.

Bisect (E. 9. 1.) the ABC, by BD, and let BD meet AC in D; through D, draw DE (E. 31. 1.) parallel to CB: E is the point which was to be found.

For, since DE is parallel to CB, the ▲ CBD= z BDE (E. 29. 1.); but (constr.) the CBD= ▲ DBE; ... < DBE=/ BDE; ... (E.6.1.) ED= EB; and since (hyp.) the C is a right 4, and that DE is parallel to CB, the CDE (E. 29. 1.) is a right; i. e. ED is 1 to AC.

33. PROBLEM. In the base of a given acute-angled triangle, to find a point, through which if a straight line be drawn perpendicular to one of the sides, the segment of the base, between that side and the point, shall be equal to the segment of the perpendicular, between the point and the other side produced.

Let ABC be the given acute-angled A: It is required, to find, in the base BC, a point through which if a perpendicular be drawn to AB, the segment of the base, between that point and the point

B, shall be equal to the segment of the perpendicular between that same point and AC produced. Draw (E. 11. 1.) from B, BY 1 to AB; bisect

(E. 9.1.) the CBY by BD, meeting AC, produced in D; through D, draw (E. 31. 1.) DE parallel to BY, and let DE cut BC in F: F is the point which was to be found.

For, sinee (constr.) the ABY is a right 2, and that DE is parallel to BY, the LE

(E. 29. 1.) is, also, a right

2 BDF; but (constr.) the DBF

the

; and the YBD YBD=4 DBF; .. BDF; ... (E. 6. 1.) FB=FD.

PROP. XXIV.

34. PROBLEM. From a given isosceles triangle to cut off a trapezium, which shall have the same base as the triangle, and shall have its three remaining sides equal to each other.

Let ABC be the given isosceles ▲: It is re

quired to cut off from it a trapezium, which, having BC for its base, shall have its three remaining sides equal to one another.

Bisect (E. 9. 1.) the ABC by BD, meeting AC in D; and through D draw (E. 31. 1.) DE parallel to CB: Then shall BE, ED, and DC, the three sides of the trapezium BEDC, be equal to one another.

For, since DE is parallel to BC, the ▲ AED= 4 ABC (E. 29. 1.), and ADE=/ ACB; but (hyp. and E. 5. 1.) ▲ ABC= 2 ACB; .'., < AED=2ADE; .. (E. 6. 1.) AE=AD; but (hyp.) AB=AC; from these equals take the equals AE and AD, there remains EBDC: Again, because DE is parallel to BC, the CBD= ▲ BDE (E. 29. 1.); but (constr.) ▲ CBD= <DBE;... the DBE BDE; ... (E. 6. 1.) EB=ED; and EB has been proved to be equal to DC; ... EB, ED and DC are equal to one another.

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